To pin down this integral we use Feynman \(\alpha \) -parametrization
(for calculation of fully massless integral with \(Q^2 = 0\)  see
Ref. [1]}). The final results can be written
in terms of the Gauss hypergeometric function
\(_2F_1(a,b,c,z)\) 
[see details of calculation in Appendix REF ].
The expression for the box integral reads
(see also Refs. [2]}, [3]}, [4]}):
\(D_0(s,u,Q^2) &=& - \frac{1}{su} \ \Biggl [\pi ^{\frac{D}{2}-2} \frac{\Gamma \Big (2-\frac{D}{2}\Big ) \,\Gamma ^2\Big (\frac{D}{2}-2\Big )}{\Gamma (D-4)}\Biggr ] \nonumber \\&\times &\Biggl [ \Big (-\frac{s}{\mu ^2}\Big )^{\frac{D}{2}-2}\ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{t}{u}\Big ) \nonumber \\&+& \Big (-\frac{u}{\mu ^2}\Big )^{\frac{D}{2}-2}\ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{t}{s}\Big ) \nonumber \\&-& \Big (-\frac{Q^2}{\mu ^2}\Big )^{\frac{D}{2}-2}\ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{Q^2 t}{s u}\Big )\Biggr ]\nonumber \\&=& - \frac{4}{su} \ \Biggl [\frac{D-3}{D-4}\Biggr ] \ \Biggl [ B_0(s) \ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{t}{u}\Big )\nonumber \\&+& B_0(u) \ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{t}{s}\Big )\nonumber \\&-& B_0(Q^2) \ _2F_1\Big (1,\frac{D}{2}-2,\frac{D}{2}-1,-\frac{Q^2 t}{s u}\Big )\Biggr ]\,.\) 
